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Q6.62:
A 0.02M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine.

Solution :

Step 1: Identifying the Chemical System

Pyridinium hydrochloride ($C_5H_5NH^+Cl^-$) is a salt formed from a weak base (pyridine, $C_5H_5N$) and a strong acid ($HCl$). In aqueous solution, it dissociates completely into $C_5H_5NH^+$ and $Cl^-$. The pyridinium ion ($C_5H_5NH^+$) undergoes hydrolysis (acidic hydrolysis) as follows:

$C_5H_5NH^+ (aq) + H_2O (l) \rightleftharpoons C_5H_5N (aq) + H_3O^+ (aq)$

Step 2: Determining $[H_3O^+]$ from pH

Given the $pH = 3.44$, we calculate the concentration of hydronium ions using the inverse logarithmic relationship [Per the definition of pH]:

$[H_3O^+] = 10^{-pH} = 10^{-3.44}$

$[H_3O^+] = 3.63 \times 10^{-4} \, M$

Step 3: Calculating the Hydrolysis Constant ($K_h$)

For the hydrolysis of a salt of a weak base and strong acid, the relationship is given by [Per the Ostwald’s Dilution Law derivation for hydrolysis]:

$K_h = \frac{[C_5H_5N][H_3O^+]}{[C_5H_5NH^+]}$

Assuming the degree of hydrolysis is small, the concentration of the pyridinium ion at equilibrium remains approximately $C = 0.02 \, M$. Since $[C_5H_5N] = [H_3O^+]$, the equation simplifies to:

$K_h = \frac{[H_3O^+]^2}{C}$

$K_h = \frac{(3.63 \times 10^{-4})^2}{0.02}$

$K_h = \frac{1.317 \times 10^{-7}}{0.02} = 6.585 \times 10^{-6}$

Step 4: Relating $K_h$ to the Ionization Constant ($K_b$)

The relationship between the hydrolysis constant of the salt and the ionization constant of the weak base is defined by [Per the relationship $K_w = K_b \times K_h$]:

$K_b = \frac{K_w}{K_h}$

Where $K_w = 1.0 \times 10^{-14}$ at $25^\circ C$.

$K_b = \frac{1.0 \times 10^{-14}}{6.585 \times 10^{-6}}$

$K_b = 1.518 \times 10^{-9}$

Hydrolysis Equilibrium Process Salt Hydrolysis Base Ionization $K_w$

Final Solution: The ionization constant of pyridine ($K_b$) is $1.52 \times 10^{-9}$.


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